Question

A stationary person observes that rain is falling vertically down at $30\text{km/h}$. A cyclist is moving up on an inclined plane making an angle ${30}^{\circ }$ with horizontal at $10\text{km/h}$. In which direction should the cyclist hold his umbrella to prevent himself from the rain ?

• A. At an angle ${\tan }^{-1}\left(\frac{3\sqrt{3}}{5}\right)$ with the inclined plane.
• B. At an angle ${\tan }^{-1}\left(\frac{3\sqrt{3}}{5}\right)$ with the horizontal.
• C. At an angle ${\tan }^{-1}\left(\frac{\sqrt{3}}{7}\right)$ with the inclined plane.
• D. At an angle ${\tan }^{-1}\left(\frac{\sqrt{3}}{7}\right)$ with the vertical.

Answer 1

Answer: (A), (D)

The correct options are
A At an angle ${\tan }^{-1}\left(\frac{3\sqrt{3}}{5}\right)$ with the inclined plane.
D At an angle ${\tan }^{-1}\left(\frac{\sqrt{3}}{7}\right)$ with the vertical.


${\overrightarrow{v}}_r=-30\hat{j},{\overrightarrow{v}}_m=\frac{10\sqrt{3}}{2}\hat{i}+\frac{10}{2}\hat{j}$
${\overrightarrow{v}}_{rm}={\overrightarrow{v}}_r-{\overrightarrow{v}}_m=-5\sqrt{3}\hat{i}-35\hat{j}$
$\tan \theta =\frac{5\sqrt{3}}{35}=\frac{\sqrt{3}}{7}\Rightarrow \theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{7}\right)$ with vertical

Angle of $v_{rm}$ with inclined plane is $60-\theta =\beta$
$\tan \beta =\tan (60-\theta )=\frac{\tan {60}^{\circ }-\tan \theta }{1+\tan {60}^{\circ }\times \tan \theta }$
$=\frac{\sqrt{3}-\frac{\sqrt{3}}{7}}{1+\sqrt{3}\times \frac{\sqrt{3}}{7}}=\frac{3\sqrt{3}}{5}$
$\Rightarrow \beta ={\tan }^{-1}\left(\frac{3\sqrt{3}}{5}\right)$